Transcript CHAPTER 14 : THE LAW OF GRAVITY 14.1) Newton’s Law of Universal Gravitation • Newton’s law of universal gravitation = every particle.

CHAPTER 14 : THE LAW OF GRAVITY
14.1) Newton’s Law of Universal Gravitation
• Newton’s law of universal gravitation = every particle in the
Universe attracts every other particle with a force that is
directly proportional to the product of their masses and
inversely proportional to the square of the distance between
them.
• If the particles have masses m1 and m2 and are separated by a
distance r, the magnitude of this gravitational force is :
m1m 2
Fg  G 2
r
(14.1)
The law of gravity
where G is a constant, called the universal gravitational
constant.
G  6.6731011 N  m2 / kg2
(14.2)
• The form of the force law given by Equation (14.1) = inversesquare law, because the magnitude of the force varies as the
inverse square of the separation of the particles.
• The force in vector form - by defining a unit vector
(Figure (14.1)
F12
F21
rˆ12 .
m2
r
rˆ12
m1
Figure (14.1)
• Because this unit vector is directed from particle 1 to particle
2, the force exerted by particle 1 on particle 2 is :
F12  G
m1m 2
rˆ12
2
r
(14.3)
where the minus sign indicates that particle 2 is attracted to
particle 1, and hence the force must be directed toward
particle 1.
• Newton’s third law - the force exerted by particle 2 on
particle 1 (F21), is equal in magnitude to F12 and in the
opposite direction (form an action-reaction pair, and
F21 = - F12 .
Properties of the gravitational force
• The gravitational force is a field force that always exists
between two particles, regardless of the medium that
separates them.
• Because the force varies as the inverse square of the distance
between the particles, it decreases rapidly with increasing
separation.
• The gravitational force exerted by a finite-size, spherically
symmetric mass distribution on a particle outside the
distribution is the same as if the entire mass of the
distribution were concentrated at the center.
• For example - the force exerted by the Earth on a particle of
mass m near the Earth’s surface has the magnitude :
Fg  G
MEm
R 2E
(14.4)
where ME is the Earth’s mass and RE its radius.
• The force is directed toward the center of the Earth.
Example (14.1) : Billiards, Anyone?
Three 0.300-kg billiard balls are placed on a table at the
corners of a right triangle (Figure (14.4)). Calculate the
gravitational force on the cue ball (designated m1) resulting
from the other two balls.
m2
0.400 m
0.500 m
y
F
F21
F31
x
m1
0.300 m
Figure (14.4)
m3
14.3) Free-Fall Acceleration and The Gravitational Force
• mg = the weight of an object of mass m, g is the magnitude
of the free-fall acceleration.
• Because the force acting on a freely falling object of mass m
near the Earth’s surface is given by Equation (14.4), we can
equate mg to this force to obtain :
mg  G
gG
ME
R 2E
MEm
R 2E
(14.5)
Free-fall acceleration near
the Earth’s surface
• Consider an object of mass m located a distance h abouve
the Earth’s surface or a distance r from the Earth’s center,
where r = RE + h.
• The magnitude of the gravitational force acting on this object
is :
Fg  G
ME m
ME m

G
r2
R E  h 2
• The gravitational force acting on the object at this position is
also Fg = mg’, where g’ is the value of the free-fall
acceleration at the altitude h.
• Substituting this expression for Fg into the last equation
shows that g’ is :
g' 
GM E
GM E

r2
R E  h 2
(14.6)
Variation of g
with altitude
• It follows that g’ decreases with increasing altitude.
• Because the weight of a body is mg’, we see that r  , its
weight approaches zero.
Example (14.2) : Variation of g with Altitude h
The Internation Space Station is designed to operate at an
altitude of 350 km. When completed, it will have a weight
(measured at the Earth’s surface) of 4.22 x 106 N. What is
its weight when in orbit?
Exampel (14.3) : The Density of the Earth
Using the fact that g = 9.80 m/s2 at the Earth’s surface, find
the average density of the Earth.
14.4) Kepler’s Law
• Kepler’s analysis first showed that the concept of circular
orbits around the Sun had to be abandoned.
• The orbit of Mars could be accurately described by an ellipse.
• Figure (14.5) - shows the geometric description of an ellipse.
• The longest dimension = the major axis and is of length 2a,
where a is the semimajor axis.
• The shortest dimension = the minor axis, of length 2b, where
b is the semiminor axis.
• On the other side of the center is a focal point , a distance c
from the center, where a2 = b2 + c2.
• The Sun is located at one of the focal points of Mar’s orbit.
• Three statements known as Kepler’s laws :
1) All planets move in elliptical orbits with the Sun at one
focal point.
2) The radius vector drawn from the Sun to a planet sweeps
out equal areas in equal time intervals.
3) The square of the orbital period of any planet is
proportional to the cube of the semimajor axis of the elliptical
orbit.
a
c
b
Figure (14.5)
F1
F2
14.5) The Law of Gravity and The motion of Planets
• The gravitational force is proportional to the inverse square
of the separation between the two interacting vodies.
• Compared the acceleration of the Moon in tis orbit with the
acceleration of an object falling near the Earth’s surface,
such as the legendary apple (Figure (14.6)).
• Assuming that both accelerations had the same cause namely, the graviational attraction of the Earth - Newton
used the inverse-square law to reason that the acceleration
of the Moon toward the Earth (centripetal acceleration)
should be proportional to 1 / rM2, where rM is the distance
between the centers of the Earth and the Moon.
• The acceleration of the apple toward the Earth should be
proportional to 1 / RE2 , where RE is the radius of the
Earth, or the distance between the centers of the Earth and
the apple.
• Using the values rM = 3.84 x 108 m and RE = 6.37 x 106 m,
Newton predicted that the ratio of the Moon’s acceleration
aM to the apple’s acceleration g would be :
aM
g
2

1 / rM 

1 / R E 2
2
2
 R E   6.37 106 m 
4
  

 

2
.
75

10
8

 rM   3.8410 m 
• Therefore, the centripetal acceleration of the Moon is :



a M  2.75104 9.80m / s2  2.70103 m / s2
• Newton also calculated the centripetal acceleration of the
Moon from a knowledge of its mean distance from the Earth
and its orbital period, T = 27.32 days = 2.36 x 106 s.
• In a time T, the Moon travels a distance 2rM , which equals
the circumference of its orvit.
• Therefore, its orbital speed is 2rM / T and its centripetal
acceleration is :

v 2 2rM / T 
4 2 rM 42 3.84108 m
aM 



2
6 2
rM
rM
T
2.3610 s
2



9.80m / s 2
 2.7210 m / s 
602
3
2
Because the Moon is roughly 60 Earth radii away, the
gravitational acceleration at that distance should be about
1 / 602 of its value at the Earth’s surface.
Kepler’s Third Law
• Kepler’s third law can be predicted from the inverse-square
law for circular orbit.
• Consider a planet of mass Mp moving around the Sun of mass
MS in a circular orbit (Figure (14.7)).
• Because the gravitational force exerted by the Sun on the
planet is a radially directed force that keeps the planet
moving in a circle, we can apply Newton’s second law
(F = ma) to the planet :
GM S M p
r
2

Mp v2
r
v
Mp
r
Figure (14.7)
MS
• Because the orbital speed v of the planet is simply 2r / T,
where T is its period of revolution, the preceding expression
2
becomes :
GM S 2r / T 

2
r
r
 4 2  3
r  K Sr 3
T  
 GM S 
2
(14.7)
Kepler’s third law
where KS is a constant given by :
42
KS 
 2.971019 s 2 / m3
GM S
The law is also valid for elliptical orbits - replace r with the
length of the semimajor axis a.
• The constant of proportionality KS is independent of the
mass of the planet - Equation (14.7) valid for any planet.
Example (14.4) : The Mass of the Sun
Calculate the mass of the Sun using the fact that the period of
the Earth’s orbit around the Sun is 3.156 x 107 s and its distance
from the Sun is 1.496 x 1011 m.
Kepler’s Second Law and Conservation of Angular
Momentum
• Consider a planet of mass Mp moving around the Sun in an
elliptical orbit (Figure (14.8)).
D
C
Sun
S
A
B
Figure (14.8)
• The gravitational force acting on the planet is always along the
radius vector, directed toward the Sun (Figure (14.9a)).
• When a force is directed toward of away from a fixed point
and is a function of r only, it is called a central force.
• The torque acting on the planet due to this force is clearly
zero; that is, because F is parallel to r,
τ  r  F  r  Frˆ  0
• From Equation (11.9) - torque equals the time rate of
change of angular momentum :  = dL / dt.
• Therefore, because the gravitational force exerted by the
Sun on a planet results in no torque on the planet, the
angular momentum L of the planet is constant :
L  r  p  r  Mp v  Mpr  v  constant
(14.8)
• Because L remains constant, the planet’s motion at any
instant is restricted to the plane formed by r and v.
• The radius vector r in Figure (14.9b) sweeps out an area
dA in a time dt.
• This area equals one-half the area |r x dr| of th
parallelogram formed by the vectors r and dr (see Section
(11.2)).
• Because the displacement of the planet in a time dt is
dr = vdt :
dA  12 r  dr  12 r  vdt 
dA
L

 constant
dt 2M p
L
dt
2M p
(14.9)
where L and Mp are both constant.
• Conclusion - the radius vector from the Sun to a planet
sweeps out equal areas in equal time intervals.
• Kepler’s second law - is a consequence of the fact that the
force of gravity is a central force, which in turn implies
that angular momentum is constant.
• Therefore, Kepler’s second law applies to any situation
involving a central force, whether inverse-square or not.
Example (14.5) : Motion in an Elliptical Orbit
A satellite of mass m moves in an elliptical orbit around the
Earth (Figure 14.10)). The minimum distance of the satellite
from the Earth is called the perigee (indicated by p in Figure
(14.10)), and the maximum distance is called the apogee
(indicated by a). If the speed of the satellite at p is vp , what is
tis speed at a?
14.6) The Gravitational Field
• Describing interactions between objects that are not in contact.
• Uses the concept of a gravitational field that exists at every
point in space.
• When a particle of mass m is placed at a point where the
gravitational field is g, the particle experiences a force
Fg = mg.
• In other words, the field exerts a force on the particle.
• Hence, the gravitational field g is defined as :
g
Fg
m
(14.10)
Gravitational field
• The gravitational field at a point in space equals the
gravitational force experienced by a test particle placed at that
point divided by the mass of the test particle.
• The presence of the test particle is not necessary for the field to
exist – the Earth creates the gravitational field.
• We call the object creating the field the source particle.
• The presence of the field can be detected, and its strength can
be measured – by placing a test particle in the field and noting
the force exerted on it.
How the field concept works?
• Consider an object of mass m near the Earth’s surface.
• Because the gravitational force acting on the object has a
magnitude GMEm / r2 (Eq. (14.4)), the field g at a distance r
from the center of the Earth is :
g
Fg
m

GM E
rˆ
2
r
(14.11)
where rˆ is a unit vector pointing radially outward from the
Earth and the minus sign indicates that the field points
toward the center of the Earth (Figure (14.11a)).
• The field vectors at different points surrounding the Earth
vary in both direction and magnitude.
• In small region near the Earth’s surface, the downward field
g is approximately constant and uniform (Figure (14.11b)).
• Equation (14.11) – valid at all points outside the Earth’s
surface, assuming that the Earth is spherical.
• At the Earth’s surface, where r = RE , g has a magnitude of
9.80 N/kg.
14.7) Gravitational Potential Energy
• Verify that the gravitational force is conservative – A force is
conservative if the work it does on an object moving between
any two points is independent of the path taken by the object.
• The gravitational force is a central force – A central force is
any force that is directed along a radial line to a fixed center
and has a magnitude that depends only on the radial
coordinate r.
• A central force can be represented by F(r) rˆ, where rˆ is a unit
vector directed from the origin to the particle
(Figure (14.12)).
• Consider a central force acting on a particle moving along the
general path P to Q in Figure (14.12).
• The path from P to Q can be approximated by a series of
steps according to the following procedure.
1) In Figure (14.12) – draw several thin wedges, which are
shown as dashed lines.
2) The outer boundary of set of wedges is a path consisting of
short radial line segments and arcs.
3) Select the length of the radial dimension of each wedge
such that the short arc at the wedge’s wide end intersects the
actual path of the particle.
4) Approximate the actual path with a series of zigzag
movements that alternate between moving along an arc and
moving along a radial line.
• By definition, a central force is always directed along one of
the radial segments; therefore, the work done by F along any
radial segment is :
dW  F  dr  F(r)dr
• By definition, the work done by a force that is perpendicular
to the displacement is zero.
• The work done in moving along any arc is zero – because F
is perpendicular to the displacement along these segments.
• Therefore, the total work done by F is the sum of the
contrivutions along the radial segments :
W  rif F(r )dr
r
Work done by a central force
Where the subscripts i and f refer to the initial and final positions.
• Because the integrand is a function only of the radial
position, this integral depends only on the initial and final
values of r.
• Thus, the work done is the same over any path from P to Q.
• Because the work done is independent of the path and
depends only on the end points – we conclude that any
central force is conservative.
• From Equation (8.2) – the change in the gravitational
potential energy associated with a given displacement is
defined as the negative of the work done by the gravitational
force during that displacement :
U  U f  U i   rif F(r )dr
r
(14.12)
• Use this result to evaluate the gravitational potential energy
function.
• Consider a particle of mass m moving between two points P
and Q above the Earth’s surface (Figure (14.13)).
• The particle is subject to the gravitational force given by
Equation (14.1).
• This force can be expressed as :
F( r )  
GM E m
r2
The negative sign indicates that the
force is attractive.
• Substituting this expression for F(r) into Equation (14.12) –
compute the change in the gravitational potential energy
function :
rf
dr
 1
U f  U i  GM E m  2  GM E m  
r
 r  ri
rf
ri
 1 1
U f  U i  GM E m  
 rf ri 
Change in gravitational potential energy
(14.13)
• The choice of a reference point for the potential energy is
completely arbitrary.
• It is customary to choose the reference point where the
force is zero.
• Taking Ui = 0 at ri =  :
U
GM E m
r
(14.14)
Gravitational potential
energy of the Earthparticle system for
separation r  R E
• The result is not valid for particles inside the Earth, where
r < RE .
• Because of our choice of Ui , the function U is always
negative (Figure (14.14)).
• Equation (14.14) can be applied to any two particles.
• The gravitational potential energy associated with any pair
of particles of masses m1 and m2 separated by a distance r
is :
Gm 1m 2
(14.15)
U
r
•
The gravitational potential energy for any pair of particles varies as 1 / r, whereas the
force between them varies as 1 / r2.
•
The potential energy is negative because the force is attractive and we have taken the
Ui = 0 when the particle separation is infinite.
•
The force between the particle is attractive – an external agent must do positive work to
increase the separation between them.
•
The work done by the external agent produces an increase in the potential energy as the
two particles are sepated.
•
That is, U becomes less negative as r increases.
• When two particles are at rest and separated by a distance r, an
external agent has to supply an energy at least equal to
+ Gm1m2 / r in order to separate the particles to an infinite
distance.
• The absolute value of the potential energy is the binding
energy of the system.
• If the external agent supplies an energy greater than the
binding energy, the excess energy of the system will be in the
form of kinetic energy when the particles are at an infinite
separation.
Three or more particles
• The total potential energy of the system is the sum over all
pairs of particles.
• Each pair contributes a term of the form given by Equation
(14.15).
• For example, if the system contains three particles (Figure
(14.15)) :
 m1m 2 m1m3 m 2 m3 

U total  U12  U13  U 23  G


r13
r23 
 r12
Figure (14.15)
----- (14.16)
2
The absolute value of Utotal –
the work needed to separate
the particles by an infinite
distance.
r12
r23
1
r13
3